# About Newton Polygon to find a solution

Given $f(x,y)$ a polynomial of two complex variables, consider the curve $f(x,y) = 0$ and say $(0,0)$ is in this curve. Does anyone know whether one can find all solutions $x = t^m, y = \sum a_rt^r$ near $(0,0)$ by using Newton Polygon algorithm?

A side question which may be easy: how can one determine the number of branches of the curve at $(0,0)$?

-
I'm not sure that there's a Newton Polygon algorithm for determining Puisex series. The Newton polygon might be involved in some method, but it certainly doesn't store enough information to do the job on its own. However, as John Mangual points out, there are certainly algorithms for determining Puisex series. –  Will Sawin May 28 '12 at 18:37
I always wondered why the Newton polygon was named after Newton... mathoverflow.net/questions/15703/newton-and-newton-polygon –  john mangual May 28 '12 at 20:37

I think you can get the Puiseux expansion by solving $t = x^{1/m}$ and setting $y = \sum a_r x^{r/m}$. Then I'd guess there are $m$ branches if $a_1 \neq 0$.

See How to Compute a Puiseux Expansion on the arXiv. These are connected with resolutions of singularities on curves.

In the paper, they try to solve $2x^4 + x^2y + 4 xy ^2 + 4 y^3 = 0$ as a function $y(x)$ in terms of $x$:

$$y(x) = c_1 x^{\gamma_1} + c_2 x^{\gamma_1 + \gamma_2} + c_3 x^{\gamma_1 + \gamma_2 + \gamma_3 } + \dots$$ The draw the newton polygon (in ASCII format)

.....
o....
x....
xo...
xxo..
xxxo.


The possible slopes are -2, -1. So they make a guess: $y = x^2(c_1 + y_1)$ where somehow $y_1(x)$ is "smaller" so we could ignore it. $$2x^4 + c_1 x^4 + 4c_1 x^5 + 4c_1 x^6 \approx (2 + c_1)x^4 = 0$$ Let's try $c_1 = -2$. Then it's possible to write an equation in $x, y_1$ and repeat this procedure. \begin{eqnarray} x^4(2 + c_1 + y_1) + 4 x^5(c_1 + y_1)^2 + 4x^6(c_1 + y_1)^3 &=& 0 \\\\ y_1 + 16 x - 16 x y_1 + 4 x y_1^2 -32x^2+48 x^2 y_1 - 24x^2 y_1^2+ 4 x^2 y_1^3 &=& 0 \\\\ \end{eqnarray}

Looking at the Newton polygon in $x,y_1$, or equivalently by noticing $$y_1 + 16 x \approx 0$$ they conclude $y_1(x) = c_1 x + c_2 x^2 + \dots$ is a power series with only one branch.

So there are only two branches for $y(x)$. In the first step we could have ignored the last two terms, $2x^4 + x^2 y \approx 0$ or $y \approx -2 x^2$. We don't feel guilty about doing these Taylor-like approximations since the curve has a singularity at $(x,y)= (0,0)$ anyway.

-

The Newton polygon of f alone does not determine the solutions nor the number of branches. (Example: $f=x^2-y^2$ and $f=(x-y)^2-y^3$ have the same Newton polygon.) However, the Newton-Puiseux theorem shows that by the Newton algorithm one can find all solutions and in particular the number of branches, and this algorithm is based on computing the Newton polygons of a sequence of polynomials starting from f and determined by changes of variables as in John Mangual's answer. This is explained in detail in Casas-Alvero's book "Singularities of plane curves" and also in Brieskorn-Knörrer "Plane Algebraic Curves".

-